Two comments: you calculator is set to degrees instead of radians (1 radian is approximately 60 degrees, so sin(1) > sin(π/6) = 0.5), and it is more efficient to divide the interval in half instead of by ten.
I need help please with the following question.
(a) Use the Intermediate Value Theorem to show that the equation x + sinx = 1, has at
least one solution in the interval [0,1].
(b) Find an approximation of the solution with an error of at most 0.05.
I have attempted the question as follows:
Part (a)
f(x) = x + sinx -1
f(0) = 0 + sin(0) -1
= 0 + 0 -1 = -1
f(1) = 1 + sin(1) - 1
= 1 + 0.017 -1 = + 0.017
Since 0 occurs between -1 and +0.017 the IVT tells us thatthere must be a number c in [0, 1] and that f(c) = 0
Therefore there exists (at least) one solution.
Part (b):
f(0) =-1, f(0.1) = -0.9, f(0.2) = -0.8, f(0.3) = -0.7,f(0.4) = -6, f(0.5) = -0.5, f(0.6) = -0.4, f(0.7) = -0.3, f(0.8) = -0.2, f(0.9)=-0.08, f(1) = +0.02
Therefore the root lies in the interval [0.9, 1]
Therefore the midpoint (0.95)approximates the root with anerror of at most 0.05
Your comments would be welcomed.
Thanks for the reply.
I can work in degrees but not sure in radians.
Example:
f(x) = x + sinx -1
f(0) = 0 + sin(0) -1
= 0 + ? -1 = -?
f(1) = 1 + sin(1) - 1
= 1 + ? -1 = + ?
How do I work the ?(sin(0) and sin(1) in radians?
Thanks again.